How do I show that a complex number $z$ is a root of $z^6-1$? This problem would be less confusing to me if one was not subtracted from $z^6$. 
Here is the problem:
If $z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$, show that $z$ is a root of $z^6-1$.
My approach to this problem involved using de Moivre's theorem to convert $z$ to it's complex polar form. I think converting it to complex polar form first makes it easy to identify the root:

$$z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$$
$$a=cos(\frac{\pi}{3}), b=sin(\frac{\pi}{3})$$
$$r=\sqrt{cos(\frac{\pi}{3})^2+sin(\frac{\pi}{3})^2}$$
$$r=1$$
$$\theta=Tan^1(\frac{b}{a})$$
$$\theta=Tan^-1(\frac{sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})})$$
$$\theta=\frac{\pi}{6}$$
$$\therefore z=1\angle{\frac{\pi}{6}}$$

This part is the confusing bit. Because $1$ is subtracted from $z^6$, I decided to expand the whole thing and then show that $z$ is a root of $z^6-1$.
$$(cos(\frac{\pi}{3})+jsin(\frac{\pi}{3}))^6-1$$
$$(\frac{1}{2}+\frac{\sqrt{3}}{2}j)^6-1$$

Noting that $j^6=-1$...
$$(\frac{1}{2}-\frac{\sqrt{3}}{2})^6-1$$

The equation becomes...
$$(-\frac{13}{32})-1$$

Which equates to...
$$-\frac{45}{32}$$

Convert to complex polar form...
$$z^6=-\frac{45}{32}\angle{0rad}$$
$$z^6=(-\frac{45}{32}\angle{0rad})^6$$
$$z^6=(-\frac{45}{32})^6\angle{(0)}^6$$
$$z^6=(-\frac{45}{32})^6\angle{6(0+2\pi n)}$$
$$z^6=(-\frac{45}{32})^6\angle{(12\pi-11\pi n)}$$
$$z^6=(-\frac{45}{32})^6\angle{(\pi n)}$$
Noting that $z$ is raised to the first power, make $n=1$
$$z^6=(-\frac{45}{32})^6\angle{\pi (1)}$$

I get...
$$z^6=(-\frac{45}{32})^6\angle{\pi}$$

Clearly the complex polar form I got is not equal to that of $z$. Can someone lead me to the right path?
 A: Alt. hint (not using the polar form): $\;z=\dfrac{1}{2}+ i\,\dfrac{\sqrt{3}}{2}\,$, so $\,|z|=1\,$, and:
$$z^2 = \left(\dfrac{1}{2}+ i\,\dfrac{\sqrt{3}}{2}\right)^2= \dfrac{1}{4}+ 2i \cdot \dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2} + i^2 \dfrac{3}{4}=-\dfrac{1}{2}+ i\,\dfrac{\sqrt{3}}{2} = - \overline{z}$$
Multiplying with $\,z\,$ gives$\,z^3 = - z \cdot \overline z = -|z|^2 = -1\,$.
A: HINT
Note that according to JuliusL33t and rtybase suggestions by De Moivre's formula


*

*$z=\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})=e^{i\frac{\pi}{3}}\implies z^6=\cos(\frac{6\pi}{3})+i\sin(\frac{6\pi}{3})$


or also by exponential form


*

*$z=\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})=e^{i\frac{\pi}{3}}\implies z^6=e^{6\cdot i\frac{\pi}{3}}$

A: Not really sure what our OP is trying to do here; what the intended method might be.  But I'm pretty sure that
$\theta = \text{Tan}^{-1} \left ( \dfrac{\sin(\pi/3)}{\cos (\pi/3)} \right ) = \text{Tan}^{-1} ( \tan (\pi/3)) = \dfrac{\pi}{3} \ne \dfrac{\pi}{6}, \tag 1$
so it seems like the first wrong turn was made at this point.  And I can't see the logical connection 'twixt
$\left ( \dfrac{1}{2} + j \dfrac{\sqrt 3}{2} \right)^6 \tag 2$
and
$\left (\dfrac{1}{2} - \dfrac{\sqrt3}{2} \right )^6; \tag 3$
so the OP's remarks from this point on are pretty much difficult for me to decipher.
Nevertheless, we do have de Moivre's formula,
$(\cos \theta + j\sin \theta)^n = \cos (n\theta) + j \sin(n\theta); \tag 4$
and if we set
$z = \cos \dfrac{\pi}{3} + j\sin \dfrac{\pi}{3}, \tag 5$
then
$z^6 = \left ( \cos \dfrac{\pi}{3} + j\sin \dfrac{\pi}{3} \right)^6 = \cos \left (6 \dfrac{\pi} 3 \right ) + j \sin \left (6 \dfrac{\pi}{3} \right) = \cos 2 \pi + j \sin 2\pi = \cos 2 \pi = 1,  \tag 6$
whence
$z^6 - 1 = 0. \tag 7$
